The acceleration of a body of mass $m$ sliding down an inclined plane with an angle of inclination $\theta$ and a coefficient of kinetic friction $\mu$ is:

$A$ uniform chain of length $L$ hangs partly from a table and is kept in equilibrium by friction. If the maximum length of the chain that can hang without slipping is $l$,then the coefficient of friction between the table and the chain is:

$A$ body is projected up along a rough inclined plane of inclination $45^{\circ}$. The coefficient of friction is $0.5$. Then the retardation of the block is

$A$ person climbs up a conveyor belt with a constant acceleration. The speed of the belt is $\sqrt{\frac{g h}{6}}$ and the coefficient of friction is $\frac{5}{3 \sqrt{3}}$. The time taken by the person to reach from $A$ to $B$ with the maximum possible acceleration is

$A$ block of mass $0.1 \, kg$ is held against a wall by applying a horizontal force of $5 \, N$ on the block. If the coefficient of friction between the block and the wall is $0.5$,the magnitude of the frictional force acting on the block is ........ $N$.

When a body slides down from rest along a smooth inclined plane making an angle of $30^{\circ}$ with the horizontal,it takes time $T$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance,it takes time $\alpha T$,where $\alpha$ is a constant greater than $1$. The coefficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$ where $x = .....$.